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A container of negligible heat capacity contains 1kg of water. It is connected by a steel rod of length 10m and area of cross-section 10cm^(2) to a large steam chamber which is maintained at 100^(@)C. If initial temperature of water is 0^(@)C, find the time after which it beomes 50^(@)C. (Neglect heat capacity of steel rod and assume no loss of heat to surroundings) (use table 3.1, take specific heat of water =4180 J//kg.^(@)C)

Answer» <html><body><p></p>Solution :Let temperature of water at <a href="https://interviewquestions.tuteehub.com/tag/time-19467" style="font-weight:bold;" target="_blank" title="Click to know more about TIME">TIME</a> t be `T`, then thermal current at time t, <br/> `I = ((100 -T)/(R ))` <br/> This increases the temperature of water form `T to T +dT` <br/> `rArr i = (<a href="https://interviewquestions.tuteehub.com/tag/dh-943253" style="font-weight:bold;" target="_blank" title="Click to know more about DH">DH</a>)/(dt) = ms (dT)/(dt)` <br/> `rArr (100-T)/(R ) = ms(dT)/(dt) rArr overset(50)underset(0)<a href="https://interviewquestions.tuteehub.com/tag/int-11513" style="font-weight:bold;" target="_blank" title="Click to know more about INT">INT</a> (dT)/(100-T) = overset(t)underset(0)int (dT)/(Rms)` <br/> `rArr -ln ((<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)/(2)) = (t)/(Rms)` <br/> or `t = Rms <a href="https://interviewquestions.tuteehub.com/tag/ln2-1076476" style="font-weight:bold;" target="_blank" title="Click to know more about LN2">LN2</a> sec` <br/> `= (L)/(KA) ms ln2 sec` <br/> `= ((10m)(1kg)(4180J//kg-^(@)C))/(46(w//m^(@)C)xx(10xx10^(-4)m^(2)))ln2` <br/> `= (418)/(46) (0.69) xx 10^(5) = 6.27 xx 10^(5) sec = 174.16 hours`</body></html>


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